What is the solution for (dy)/(dx)=(x+y)^2-1,y(3)=4 ?

The solution for (dy)/(dx)=(x+y)^2-1,y(3)=4 is y=-7/(7x-22)-x

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(dy)/(dx)=(x+y)^2-1,y(3)=4

\frac{d y}{d x} = (x + y)^{2} - 1, y (3) = 4

y = - \frac{7}{7 x - 22} - x

Solution steps

\frac{d y}{d x} = (x + y)^{2} - 1

Solve using substitution:

y = - \frac{7}{7 x - 22} - x

y = - \frac{7}{7 x - 22} - x

\int_{0}^{1} 2 (1 + x)^{8} d x

\frac{( x + 2 )}{( x - 2 )}

x \to 3 lim (\frac{2 ^{x} - 8}{3 - x})

y = 4 x, y = 4, y = \frac{4}{9} x^{2} 4

\int_{6}^{8} \frac{62}{( x - 6 ) ^{3}} d x

What is the solution for (dy)/(dx)=(x+y)^2-1,y(3)=4 ?
The solution for (dy)/(dx)=(x+y)^2-1,y(3)=4 is y=-7/(7x-22)-x